Coffee Hour Problems of the Week (solutions)

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1 Coffee Hour Problems of the Week (solutios) Edited by Matthew McMulle Otterbei Uiversity Fall 0 Week. Proposed by Matthew McMulle. A regular hexago with area 3 is iscribed i a circle. Fid the area of a regular hexago circumscribed about the same circle. Solutio. Let r deote the radius of the circle. The iscribed hexago is comprised of six equilateral triagles with side legth r, while the circumscribed hexago is comprised of six equilateral triagles with height r. Thus, the area of the iscribed hexago is 6 r/ 3r/, ad the area of the circumscribed hexago is 6 3r/3 r. Usig the fact that the iscribed hexago has area 3, we get = 6r 3. Thus, the area of the circumscribed hexago is 6r 3/3 = /3 = 4. Week. Proposed by Matthew McMulle. A regular -go with area A is iscribed i a circle. Fid the area of a regular -go circumscribed about the same circle (as a fuctio of A ad ). Solutio. Let r deote the radius of the circle. Similar to the above solutio, we divide each -go ito cogruet isosceles triagles ad use trig to fid the side legths. We fid that the area of the iscribed polygo is r si(π/) r cos(π/), while the area of the circumscribed polygo is r ta(π/) r. Thus, i terms of A ad, the area we seek is give by A sec (π/).

2 Week 3. Proposed by Matthew McMulle. Ca you fid two real umbers, a ad b, such that a > 0 ad (ax + b) dx = (ax + b) dx = (ax + b) 3 dx? Solutio. Yes! Itegratig the above equatios gives a + b = a 3 + ab + b = a3 4 + a b + 3ab + b 3. Settig the first part equal to the last part, ad factorig, yields (b + a)(b + ab + a ) = 0. If b + a = 0, the we get a = b = 0. Thus, we eed to solve the system of quadratic equatios b + ab + a = 0 ad 6b + (6a 6)b + a 3a = 0. We ll leave off the (gory) details ad just say that the oly solutio with a > 0 is a = 3 ad b = ( 3)/. Week 4. Proposed by Matthew McMulle. Show that exists ad is betwee 0.5 ad 0.6. (( lim ) ) l Solutio. Let H = k= k. We will show that the sequece (H l ) is decreasig ad bouded below by 0.5. Showig this sequece is decreasig is equivalet to showig that l( + ) l > +, for all. Put f(x) = l(x + ) l x /(x + ). The f (x) = /((x(x + ) ) < 0, for all x > 0. Therefore, for all x > 0, f(x) > lim u f(u) = 0. I particular, we have show that l( + ) l > +, for all positive itegers. To show that our sequece is bouded below by 0.5, we use the Trapezoid Rule to estimate (/x) dx. Sice the fuctio y = /x is covex o (0, ), this will be a overestimate of the itegral. Thus, ( + ) + ( + ) + + ( 3 + ) > dx = l. x

3 I other words, for all >, H l > + >. Sice (H l ) is decreasig ad bouded below by 0.5, we have show that the limit i questio exists ad is at least 0.5. To show that the limit does ot exceed 0.6, simply ote that H l = < 0.6. Week 5. Proposed by Rya Berdt ad Matthew McMulle. Let s = k= k. Does coverge? What about = = s s? Solutio. Usig last week s problem ad the limit compariso test, we see that the first sum is equicoverget to the series = l, which diverges by the itegral test. Similarly, the secod sum is equicoverget to the series (l ), = which coverges by the itegral test. 3

4 Week 6. Proposed by Zegxiag Tog. I the followig diagram, BD = CD ad quadrilaterals ABEF ad ACGH are squares. Prove AD = F H. Solutio. Locate the figure i the plae; where, without loss of geerality, B = (0, 0), D = (, 0), ad A = (x, y), for some x, y with y > 0. The it ca be show that F = (x y, x + y) ad H = (x + y, y x + ). The Thus, AD = F H. F H = (y) + (x ) = 4(y + (x ) ) = 4 AD. Solutio. (By Stephe Sheema, Computer Sciece major.) Add the poit X to the figure so that ACXB is a parallelogram. We are give that AC = AH ad CX = AB = AF. Notice that ACX + CAB = 80 = F AH + CAB, so ACX = F AH. Thus, by SAS, triagles ACX ad HAF are cogruet; i particular, AX = F H. Sice the diagoals of a parallelogram bisect each other, D is o AX ad AD = AX = F H. Week 7. Proposed by Matthew McMulle. For ay prime p > 3, prove that 3 divides 0 p 0 p +. Ca you classify all oegative itegers such that 3 divides 0 0 +? Solutio. I our solutio, all of the cogrueces will be modulo 3. First, sice 00 = 3 77, 0 3. Next, sice x 3 + = (x + )(x x + ), we have that = (0 + )(0 0 + ), () for all oegative itegers. If is eve, () implies (0 + )(0 0 + ); 4

thus, sice 3 is prime, we caot have 0 0 + 0. Now suppose = 6k +, for some oegative iteger k. The () implies 0 (0 0 + ), ad it follows that 3 divides 0 0 +.

5 thus, sice 3 is prime, we caot have Now suppose = 6k +, for some oegative iteger k. The () implies 0 (0 0 + ), ad it follows that 3 divides Similarly, if = 6k, for some positive iteger k, () implies 0 5(0 0 + ), ad agai 3 divides Fially, suppose = 6k + 3, for some oegative iteger k. The = (0 6 ) k+ (0 6 ) k I summary, 3 divides if ad oly if is either oe more or oe less tha a multiple of 6 (otice that this icludes all primes greater tha 3). Week 8. Proposed by Zegxiag Tog. I the followig diagram, BAC = ABC ad CD bisects ACB. Show that BC = CA + AD. Solutio. Exted CA past A to the poit X such that AX = AD. Notice that DAX = π BAC = π ABC. Sice AXD is isosceles, AXD = (π DAX)/ = ABC. Also, we are give that ACD = BCD. Thus, by AAS, BCD = XCD. I particular, BC = XC = CA+AX = CA+AD. Solutio. Let α deote ABC, ad let β deote ACD = BCD. Without loss of geerality, suppose CA =. By the Law of Sies applied to ABC, si α = BC si α = BC si α cos α. Thus, BC = cos α; ad we eed to show that AD = cos α. 5

6 By the Law of Sies applied to CAD, AD si β = si(π (α + β)). () We ext ote that 3α + β = π, or β = (π 3α)/. Thus, () yields si(π/ 3α/) AD = si(π/ α/) = cos(3α/) cos(α/) cos(α + α/) = cos(α/) = cos α si α si(α/) cos(α/) = cos α si (α/) = cos α + ( si (α/)) = cos α. Week 9. Proposed by Matthew McMulle. State ad prove the geeral result illustrated by the fact that 4 = 6, 34 = 56, 334 = 556, ad 3334 = Ca you fid similar results i bases other tha 0? Solutio. Let 0. The geeral result ca be represeted by (3. {{.. 3 4) =. {{.. 5. {{ I summatio otatio, this ca be writte as ( + ) 3 0 k = + k=0 5 0 k + k=0 0 k++. Put u = k=0 0k. The we eed to show that ( + 3u) = + 5u u. This equatio has two solutios: u = 0 ad u = (0 + )/9, so we re doe. To geeralize to base b, we work backwards ad let u = k=0 bk. The u = (b + )/(b ); or, k=0 + (b )u = b + u u +. (3) 6

7 To complete the square o the left-had side, we must have b = a +, for some a. The (3) is equivalet to ( + au) = + (a )u + b + u. I other words, for ay a > ad ay 0, i base a +. (a. {{.. a (a + )) =. {{.. (a )... (a ) (a), {{ + Week 0. Proposed by Matthew McMulle. Show that arccos 5 = arcta 3. Solutio. Put u = arcta 3 (so 0 < u < π/). The ta u = 3 solvig the associated right triagle tells us that cos u = 5. Thus, cos(u) = cos u = 3 5 = 5. 3, ad Sice 0 < u < π, arccos(cos(u)) = u; i other words, arccos 5 = u = arcta 3. Week. Proposed by Matthew McMulle. (a) What is the expected umber of times you must roll a fair die to get two cosecutive sixes? (b) Your fried bets you that it will take at least 30 rolls for you to get two cosecutive sixes. Should you take this bet? Solutio. (a) Let p() be the probability that it takes rolls to get two cosecutive sixes. The p() = 0 ad p() = /36. Suppose >. If our first roll is ot a six, the we have rolls to get two cosecutive sixes. If our first roll is a six, the our ext roll caot be a six (sice > ), ad we have rolls to get two cosecutive sixes. Thus, for >, p() = p( ) + p( ). (4) 36 7

8 We eed to fid the expected value of. First, sice p() is a probability distributio, = p() =. Let x = = p(). The, usig (4), we have x = p() + 5 p( ) + 5 p( ) 6 36 = = =3 =3 ( + ) p() + 5 ( + ) p() 36 = (x + ) + (x + ) Solvig for x gives a expected value of 4 rolls. For (b), we use Mathematica to compute Pr( 30) = Pr( 9) = 9 = p() = So we are slightly more likely tha ot to require 30 or more rolls. You should ot take your fried s bet! = Week. Proposed by Matthew McMulle. Let x 0. Show that ( + x)( + x ) ( + x 3 ) ( + x ) 3. Solutio. By cacelig + x from both sides ad rearragig factors, what we are tryig to show is equivalet to ( + x)( + x 3 ) ( + x )( + x ) ( + x )( + x 3 ) ( + x ). To prove this, we will show that ( + x )( + x 4 ) ( + x ), (5) for all =,,...,. After multiplyig out, rearragig, ad factorig, (5) is equivalet to the statemet which is clearly true. x (x ) 0, Similarly, oe ca show that the stadard deviatio of this probability distributio is Moreover, we have show that equality is attaied if ad oly if x = 0 or x =. 8

9 Week 3. From The College Mathematics Joural. Show that ( ) ( ) x + 3 x 3 si si + 4x + 4x is costat for x. Solutio. Put ( ) ( ) x + 3 x 3 f(x) = si si. + 4x + 4x The oe ca show, after a bit of work, that ( 3 x f (x) = x + x + x ). + x Thus, f (x) = 0 for < x <. Sice f is cotiuous, this meas that f is costat o [, ]. Solutio. Put ( ) x + 3 u = si + 4x ( ) x 3 ad v = si. + 4x The si u = x+3, si v = x 3, cos u = 3 x, ad cos v = 3 +x. +4x +4x +4x +4x Thus, for x, we obtai, after a bit of work, 3 si(u v) = si u cos v cos u si v =. Sice u v is cotiuous, we either have u v π/3 or u v π/3 o [, ]. 3 3 Pluggig i x = 0 shows us that u v π/3 o [, ]. 9

10 Week 4. Proposed by Zegxiag Tog. I the followig diagram, circles O ad P are taget, AB is taget to both circles, BC is a diameter of circle O, ad CD is taget to circle P. Show that BC = CD. Solutio. I the diagram below, E is the poit of tagecy of circles O ad P, ad EF is taget to both circles. Sice BC is a diameter, BEC = 90. Sice AB is taget to circle O, ABC = 90. Sice F A ad F E are taget to the same circle, F A = F E; hece, F AE = F EA. Similarly, F B = F E, ad so F BE = F EB. We have 80 = F EB + F EA. Thus, AEB = 90 ad poits C, E, ad A are colliear. Now, sice triagle ABC is similar to triagle BEC ad triagle ADC is similar to triagle DEC, BC = CA CE = CD, ad we are doe! 0

11 Week 5. Purdue Uiversity Problem of the Week. What is the maximum value of a ad the miimum value of b for which ( + ) +a ( e + ) +b for every positive iteger? Solutio. The give iequality is equivalet to a l( + /) b. Put f(x) = / l( + /x) x. We have two claims: lim x f(x) = / ad f is icreasig o (0, ). Sice f is cotiuous for x > 0, provig these claims will show that the maximum value of a is f() = / l ad the miimum value of b is /. For the first claim we have (usig little-o otatio) x l( + /x) lim f(x) = lim x x l( + /x) x(/x /(x ) + o(/x )) = lim x /x + o(/x) /(x) + o(/x) = lim x /x + o(/x) =. For the secod claim, we will show that f > 0 o (0, ). After some maipulatio (most otably takig square roots ad makig the substitutio + /x x), this is equivalet to showig that l x < x /x o (, ). Put g(x) = x /x l x. The g (x) = + /x /x = ( /x) > 0 o (, ). Thus, for all x >, g(x) > g() = 0; ad therefore f > 0 o (0, ).

Solutions for May. 3 x + 7 = 4 x x +

Solutions for May. 3 x + 7 = 4 x x + Solutios for May 493. Prove that there is a atural umber with the followig characteristics: a) it is a multiple of 007; b) the first four digits i its decimal represetatio are 009; c) the last four digits